How to find the rotation angle and axis of rotation of linear transformation? I need some help with this problem:

We know that $T(x_1,x_2,x_3)=(x_2,x_3,x_1)$. We suspect that it is a rotation matrix, to be sure, we need to determine the rotation axis and the rotation angle.

I first tried to see what the linear transformation did to the basis vectors.
$$T(1,0,0)=(0,0,1)$$ $$T(0,1,0)=(1,0,0)$$ $$T(0,0,1)=(0,1,0)$$ With this informatio, I wrote the matrix that represents the linear transformation: $$A=\left[\begin{matrix} 
0&1&0\\0&0&1\\1&0&0 
\end{matrix}\right]$$
After that, I don't know how to determine the angle and the axis, can you help me?
 A: A vector on the axis of the rotation matrix $A$ will satisfy $Av=v$, that is $(A-I)v=0$. In this example,
$$A-I=\pmatrix{-1&1&0\\0&-1&1\\1&0&-1}.$$
and $(A-I)\pmatrix{x\\y\\z}=\pmatrix{0\\0\\0}$ iff $x=y=z$, so the axis is
the line joining the origin to $(1,1,1)$.
In this example, $A^3=I$, so the angle of rotation is a third of a turn: $\frac{2\pi}3$.
More generally if $\theta$ is the rotation angle of a rotation matrix $A$,
then the trace of $A$ is $1+2\cos\theta$, so the angle can easily be found from this.
A: Let me try it in a simple way. First determine it is a rotation matrix or not. To be a rotation matrix the matrix should be orthogonal and the determinant of the matrix should be +1 (if it is orthogonal and determinant is -1, then it is not a rotation matrix). And for this matrix $\begin{bmatrix} 0&1&0\\0&0&1\\1&0&0 \end{bmatrix}$ we can easily confirm it is orthogonal and determinant is +1.
It is probably clear that a vector on the axis of the rotation matrix A will not change direction (as it is on the axis of rotation) (as mentioned by Lord Shark the Unknown), So treating this vector as $(x,y,z)$, we see that 
$\begin{bmatrix} 0&1&0\\0&0&1\\1&0&0 \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}x\\y\\z\end{bmatrix}$
Now we solve these linear equations.
The first equation gives $0.x+1.y+0.z = x$, so $y=x$
The second equation gives $0.x+0.y+1.z = y$, so $z=y$
The third equation gives $1.x+0.y+0.z = z$, so $x=z$
That gives the axis of rotation x=y=z, you can take this as along the line of vector (1,1,1) or (-1,-1,-1).
Now regarding the angle of rotation it may be little complex.The angle of rotation in this case is the rotation of the plane which is perpendicular to rotation axis. To find out the magnitude and the direction we need to see a vector on this plane, how much it was rotated and in which direction. Now the direction is relative, based on from which side of the plane you are looking at. 
From x=y=z, we can form vector (1,1,1), (-1,-1,-1), Now any vector (x,y,z) perpendicular to these vector will lie on the plane perpendicular to these vector. To find out arbitrarily we set the dot product to zero i.e.
$x.1+y.1+z.1=0$ or $z=-x-y$, we also get from condition of rotation the new vector will be (y,z,x) . So putting arbitrarily values for x,y we get vector A (1,2,-3), and after rotation vector B(2,-3,1). Let us again use the dot product to find out the angle between these vectors
$AB\cos\theta = 2-6-3$ or , $\sqrt{14}\sqrt{14}\cos\theta = -7$, or $\cos\theta =-1/2$ and we get $\theta =\pm 2\dfrac{\pi}{3}$. Obviously $\theta$ can be both positive and negative based on from which side we are looking at the plane.
To figure out this we use cross product. Let the cross product (the vector) is C. We find out its component,
$\begin{bmatrix}1&2&-3\\2&-3&1\\i&j&k\end{bmatrix}$
$C_x =\begin{bmatrix}2&-3\\-3&1\end{bmatrix} = -7$, 
$C_y =-\begin{bmatrix}1&-3\\2&1\end{bmatrix} = -7$, 
$C_z =\begin{bmatrix}1&2\\2&-3\end{bmatrix} = -7$, 
So C(-7,-7,-7), we find out angle from the relation $C=AB\sin\theta$ or, $7\sqrt{3}=\sqrt{14}\sqrt{14}\sin\theta$, $\sin\theta = \dfrac{\sqrt{3}}{2}$ 
so $\theta = \dfrac{\pi}{3}, 2\dfrac{\pi}{3}$
Now what we found out from cross product is that looking at the plane from cross product C(-7,-7,-7), vector B is in the positive direction of, i.e. anticlockwise from vector A by positive angles $\dfrac{\pi}{3}, 2\dfrac{\pi}{3}$ (to get the same cross product, these two different angles are possible).
Comparing this with the result of dot product we see that the required angle is $+2\dfrac{\pi}{3}$, Now this angle, what we see is from the side of C(-7,-7,-7) or, (-1,-1,-1).   
A: From this link: https://arxiv.org/abs/1404.6055 it says,
if a square matrix is a rotation, you can find the rotation angle from its complex eigenvalues, and the rotation axis from its eigen vectors corresponding to its real eigen value.
For this specific case, the real eigen value 1's corresponding eigen vector is $(1,1,1)^T$, which is the rotation axis (center); the two complex eigen values are $\dfrac{1}{2} \left(-1 \pm i \sqrt{3}\right)$ , rotation angle is $2k\pi\pm\dfrac{2\pi}3$; the symbol of rotation angle is left- right- hand coordinate system choice dependent.
